Mathematics • Year 9 • Unit 4 • Lesson 14

Two-Stage Probability in the Real World

Use tree diagrams and the multiplication rule on real situations, lolly bags, school traffic lights, lucky-dip prizes, rainy-day umbrellas and Punnett-square genetics. Then explain "with vs without replacement" in your own words.

Apply · Real-World Maths

1. Word problems

Each problem uses the multiplication rule, a tree diagram, or the with/without replacement contrast from Lesson 14. Show your working, a single final answer with no working only earns half marks.

1.1, Lolly bag. A canteen lolly bag has 6 red and 4 green lollies. A student takes two at random without replacement.

(a) Draw a tree diagram for the two draws.
(b) Find P(both red) and P(one of each colour). 4 marks

Stuck? After the first draw the bag has 9 lollies, not 10. The second-stage fractions depend on what came out first.

1.2, Traffic lights on the way to school. A student passes two independent sets of lights. P(green at light 1) = 0.6 and P(green at light 2) = 0.7.

(a) Find P(both lights are green).
(b) Find P(at least one light is red) using the complement rule. 4 marks

Stuck? "At least one red" is the opposite of "both green". Subtract from 1.

1.3, Lucky-dip prizes. A school fair lucky dip has 20 envelopes: 4 small prizes, 6 medium prizes and 10 blank. You and a friend each draw one envelope (yours first, without replacement).

(a) Find P(you both get a prize of any size).
(b) Find P(you both get blanks). 4 marks

Stuck? "Prize of any size" = small + medium = 10 envelopes. After you draw a prize there are 19 envelopes left with 9 prize envelopes still inside.

1.4, Weather and umbrella. On any given day in autumn, P(it rains) = 0.4. Separately, P(a student brings an umbrella to school) = 0.5 (the two events are independent, whether it rains does not depend on whether the student decided to bring one).

(a) Draw a tree diagram with two stages: rain/no rain then umbrella/no umbrella.
(b) Find P(it rains AND the student has an umbrella).
(c) Find P(it rains AND the student has NO umbrella). 4 marks

Stuck? The two stages are independent, so P(rain) stays 0.4 and P(umbrella) stays 0.5 on every branch. Multiply along each branch.

1.5, Punnett-square genetics. Both parents carry the gene combination Aa for a trait. Each parent independently passes on A or a with equal probability. (This is exactly a two-stage tree where each branch is 1/2.)

(a) List the four equally likely combinations the child can inherit.
(b) Find P(child inherits AA), P(child inherits aa) and P(child inherits Aa or aA). 4 marks

Stuck? Revisit lesson § "Real-World Anchor, Genetics and Medicine", the Punnett square is a two-stage tree where each parent contributes one letter.

2. Explain your thinking

This question is about communication, not just answers. Use full sentences. 4 marks

2.1 A classmate says: "For two events in a row, P(A and B) is always P(A) × P(B)." Using a 4-red / 2-blue bag and the question "what is P(both red)?" as your example, explain in your own words (i) when their formula works (with replacement), (ii) why it does not work without replacement, and (iii) the correct calculation in each case. Mention "independent" and "dependent" somewhere in your answer.

Stuck? Revisit lesson § "Misconceptions", treating without replacement as with replacement is the most common error in this topic.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers, Do not peek before attempting

1.1, Lolly bag (6R/4G, without replacement)

Tree (stage 1 → stage 2):

        6/10 R --- 5/9 R   (RR  30/90)
       /          \--- 4/9 G   (RG  24/90)
 start
       \
        4/10 G --- 6/9 R   (GR  24/90)
                   \--- 3/9 G   (GG  12/90)

P(both red) = 6/10 × 5/9 = 30/90 = 1/3.
P(one of each) = P(RG) + P(GR) = 24/90 + 24/90 = 48/90 = 8/15.

1.2, Traffic lights

(a) P(both green) = 0.6 × 0.7 = 0.42.
(b) P(at least one red) = 1 − P(both green) = 1 − 0.42 = 0.58.

1.3, Lucky dip

Prizes = 4 + 6 = 10 envelopes; blanks = 10 envelopes; total = 20.
(a) P(both prize) = 10/20 × 9/19 = 90/380 = 9/38.
(b) P(both blank) = 10/20 × 9/19 = 9/38 (same, by symmetry, prizes and blanks each total 10).

1.4, Weather and umbrella

Tree (independent branches, the second-stage probabilities are the same on every branch):

        0.4  Rain --- 0.5 Umbrella   (Rain+Umb   0.20)
       /              \--- 0.5 No umb  (Rain+NoUmb 0.20)
 start
       \
        0.6  No rain --- 0.5 Umbrella  (NoRain+Umb  0.30)
                         \--- 0.5 No umb (NoRain+NoUmb 0.30)

(b) P(rain AND umbrella) = 0.4 × 0.5 = 0.20.
(c) P(rain AND no umbrella) = 0.4 × 0.5 = 0.20.
Check: all four branches sum to 0.20 + 0.20 + 0.30 + 0.30 = 1.00 ✓

1.5, Punnett (Aa × Aa)

(a) Four equally likely combinations: AA, Aa, aA, aa (each prob 1/4).
(b) P(AA) = 1/4. P(aa) = 1/4. P(Aa or aA) = 2 × 1/4 = 1/2. The totals correctly add to 1.

2.1, Explain your thinking (sample response)

The formula P(A and B) = P(A) × P(B) only works when the two events are independent. With a 4-red / 2-blue bag, drawing with replacement makes the two draws independent because the bag is restored between draws: P(both red) = 4/6 × 4/6 = 16/36 = 4/9. Drawing without replacement makes the two draws dependent after a red is removed, only 3 red and 2 blue remain, so the second probability changes: P(both red) = 4/6 × 3/5 = 12/30 = 2/5. The structure of the formula stays the same (multiply along the branch), but the numbers on the second branch change because the bag itself has changed.

Marking: 1 mark for correct "with replacement" calculation; 1 mark for correct "without replacement" calculation; 1 mark for explicit reference to independent/dependent; 1 mark for a clear sentence-based contrast.